9 Boundedness

Highlights

Definition of a bounded subset of a metric space
Equivalent conditions to boundedness
Boundedness in \(\R\) is equivalent to “bounded above and below”

9.1 Preparation

Theorem 9.1 Let \(X\) be a metric space and let \(S\subseteq X.\) Then the following are equivalent.

For any \(a\in X,\) there exists \(r>0\) such that \(S\subseteq \bee_r(a).\)
There exists \(a\in X\) and \(r>0\) such that \(S\subseteq \bee_r(a).\)

In this case, we say \(S\) is bounded.

Desmos Applet 9.1: An illustration of Condition 1 of the boundedness condition. Wherever the point \(a\) is moved, there is a large enough \(r\) so that the set \(S\) lives inside the open ball.

You will prove Theorem 9.1 in Exercise 9.4 below.

Note

Theorem 9.1 is both a theorem and a definition. Once we prove the conditions are equivalent, we can refer to any one of them as “the” definition of bounded.

Exercise 9.2

If a problem asks you to prove that a set is bounded, which of the two conditions from Theorem 9.1 would you choose to work with, and why?
If a problem tells you to assume that a set is bounded and you need to apply boundedness in order to prove something else, which of the two conditions from Theorem 9.1 would you choose to work with, and why?

Exercise 9.3 Negate each of the two equivalent characterizations of boundedness.

A subset of a metric space that is not bounded is called unbounded.

9.2 Properties of Boundedness

We start with proving that the two definitions of boundedness are in fact equivalent.

Exercise 9.4 Prove Theorem 9.1.

Hint

One of the two implications should be very quick to prove!
Note that \(a\) might lie outside of \(S\) for either of the two conditions.

Exercise 9.5 Let \(S=\{(n,n)\in\R^2\st n\in\N\}.\) Draw a picture of \(S\) and give a careful proof that \(S\) is unbounded.

Exercise 9.6 Suppose \(S_1\) and \(S_2\) are bounded subsets of a metric space \(X.\) Prove that \(S_1\cup S_2\) is bounded.

9.3 Boundedness in \(\R\)

As ever, what we really care about is how this concept applies to \(\R\) and how the concepts defined thus far interact. We’ve already defined “bounded above” and “bounded below”. The next theorem will connect these definitions to our new more general definition of “bounded” for metric spaces.

Theorem 9.7 Suppose \(S\subseteq\R.\) Then the following are equivalent.

\(S\) is bounded.
\(S\) has an upper bound and a lower bound.
There exists \(K>0\) such that for all \(x\in S,\) \(|x|<K.\)

Exercise 9.8 Prove Theorem 9.7.

Desmos Applet 9.2: An illustration of the three equivalent conditions for the boundedness of \(S\) in \(\R\): an open ball \(\bee_r(a),\) the set of \(x\) with \(|x|<K,\) and an upper and lower bound, \(u\) and \(l,\) on \(S.\) A hint for Exercise 9.8 is to think about how to construct the variables for one construction from another.

Exercise 9.9 Prove that the complement of a bounded subset of \(\R\) is unbounded.

Review Questions

What does it mean for a subset of a metric space to be bounded? Unbounded?
With Theorem 9.1 and Theorem 9.7, we have four equivalent ways of describing boundedness in \(\R\); what are they?
Building on Review Question 1 from Chapter 6, think about what theorems or facts relate the major concepts we’ve encountered thus far.

Figure 9.1